package algorithm.problems.breadth_first_search;
import java.util.*;
/**
 * Created by gouthamvidyapradhan on 23/06/2018.
 *
 * You are asked to cut off trees in a forest for a golf event. The forest is represented as a non-negative 2D map,
 * in this map:

 0 represents the obstacle can't be reached.
 1 represents the ground can be walked through.
 The place with number bigger than 1 represents a tree can be walked through, and this positive number represents the
 tree's height.
 You are asked to cut off all the trees in this forest in the order of tree's height - always cut off the tree with
 lowest height first. And after cutting, the original place has the tree will become a grass (value 1).

 You will start from the point (0, 0) and you should output the minimum steps you need to walk to cut off all the
 trees.  If you can't cut off all the trees, output -1 in that situation.

 You are guaranteed that no two trees have the same height and there is at least one tree needs to be cut off.

 Example 1:
 Input:
 [
 [1,2,3],
 [0,0,4],
 [7,6,5]
 ]
 Output: 6
 Example 2:
 Input:
 [
 [1,2,3],
 [0,0,0],
 [7,6,5]
 ]
 Output: -1
 Example 3:
 Input:
 [
 [2,3,4],
 [0,0,5],
 [8,7,6]
 ]
 Output: 6
 Explanation: You started from the point (0,0) and you can cut off the tree in (0,0) directly without walking.
 Hint: size of the given matrix will not exceed 50x50.


 Solution: O(N x M) ^ 2: Bfs to each height starting from 1 and calculate the total sum of distance.
 */
public class CutOffTreesForGolfEvent {

    public static void main(String[] args) throws Exception{

    }

    private static final int[] R = {0, 0, 1, -1};
    private static final int[] C = {1, -1, 0, 0};

    static class Cell implements Comparable<Cell>{
        int r, c;
        int distance;
        int height;
        Cell(int r, int c){
            this.r = r;
            this.c = c;
        }
        @Override
        public int compareTo(Cell o) {
            return Integer.compare(this.height, o.height);
        }
    }

    public int cutOffTree(List<List<Integer>> forest) {
        int distance = 0;
        List<Cell> trees = new ArrayList<>();
        for(int i = 0; i < forest.size(); i ++){
            for(int j = 0; j < forest.get(0).size(); j ++) {
                if(forest.get(i).get(j) > 1){
                    Cell cell = new Cell(i, j);
                    cell.height = forest.get(i).get(j);
                    trees.add(cell);
                }
            }
        }
        Collections.sort(trees);
        int sR = 0, sC = 0;
        for(Cell t : trees){
            int dist = bfs(forest, t.height, sR, sC);
            if(dist == -1) return -1;
            else distance += dist;
            sR = t.r;
            sC = t.c;
        }
        return distance;
    }

    private int bfs(List<List<Integer>> forest, int target, int sR, int sC){
        if(forest.get(sR).get(sC) == target) {
            forest.get(sR).set(sC, 1);
            return 0;
        }
        Cell start = new Cell(sR, sC);
        start.distance = 0;
        Queue<Cell> queue = new ArrayDeque<>();
        queue.add(start);
        boolean[][] done = new boolean[forest.size()][forest.get(0).size()];
        done[sR][sC] = true;
        while(!queue.isEmpty()){
            Cell cell = queue.poll();
            for(int i = 0; i < 4; i ++){
                int newR = cell.r + R[i];
                int newC = cell.c + C[i];
                Cell newCell = new Cell(newR, newC);
                if(newR >= 0 && newR < forest.size() && newC >= 0 && newC < forest.get(0).size() &&
                        forest.get(newR).get(newC) != 0 && !done[newCell.r][newCell.c]) {
                    newCell.distance = cell.distance + 1;
                    if(forest.get(newR).get(newC) == target){
                        forest.get(newR).set(newC, 1);
                        return newCell.distance;
                    }
                    done[newCell.r][newCell.c] = true;
                    queue.offer(newCell);
                }
            }
        }
        return -1;
    }
}
